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3 years ago

If 390 words were typed in 6 min how many words were typed in 10 min

Mathematics

2 answers:

ololo11 [35]3 years ago

8 0

650 words were typed in ten minutes

umka21 [38]3 years ago

3 0

390/6=65
65x10+650
650 is your answer
hope this helps

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What are the factors of 18 24 36

Natalka [10]

Factors of 18: 1, 2, 3, 6, 9, 18

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

4 0

3 years ago

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Can anyone help me with this problem???

jasenka [17]

Note that x is the hypotenuse of a right triangle. Applying the Pyth. Thm.,

x^2 = 2^2 + 6^2, or x^2 = 4 + 36 = 40.

Thus, the hypotenuse is +sqrt(40), or 2sqrt(10).

8 0

3 years ago

Find the LCM of 315,420,525 using prime factorisation method with slove it

Vedmedyk [2.9K]

$\boxed{\sf \:\begin{cases}\\\begin{gathered} {\underline{{ \sf {\blue{\Large\bf \underbrace{ {\rm {\purple{\begin{array}{r | l}\large\rm{\red{ 5}}&\underline{315,420,525}\\\large\rm{\red{ 3}}& \underline{63,84,105}\\\large\rm{\red{3}}&\underline{21,28,35}\\\large\rm{\red{ 7}}&\underline{7,28,35}\\\large\rm{\red{2}}&\underline{1,4,5}\\\large\rm{\red{2}}&\underline{1,2,5}\\\large\rm{\red{5}}&\underline{1,1,5}\\&1,1,1\end{array}}}}} }}}}}\end{gathered}\\\end{cases}}$

LCM = 5 × 3 × 3 × 7 × 2 × 2 × 5

LCM = 6300

<u>Hence, the LCM of 315 , 420 , 525 is 6300</u>

5 0

3 years ago

Read 2 more answers

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago

For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th

timurjin [86]

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

$(a,b)=\left \{ x\in\Re : a$

2. Functions

1) $\Re \rightarrow \Re \\ f(x)=x$

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2) $\Re\geq0\rightarrow\Re , f(x)=x+1\\$

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3) $f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)$

The cosine function is not injective, bijective nor surjective.

4) $f:\Re\rightarrow\Re \:f(x)=ex$

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5) Quite unclear format

$6) \:f:\Re\rightarrow(0,\infty), f(x) =ex$

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

$7) f:\Re\geq 0 \rightarrow \Re\geq0, f(x) =x^{4}$

Not injective nor surjective therefore not bijective too.

$8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}$

$f(-1)=1, f(2)=4, f(-3)=9$

Injective (one to one), Surjective, and Bijective.

$10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.$

Surjective.

$11.f:R\geq 0[-1,1], f(x) = 0\\$

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

Surjective

7 0

3 years ago